Rails PG :: UndefinedTable：错误：缺少表的FROM子句条目

Question

我有模特

class Offer < ActiveRecord::Base
  belongs_to :agency
end

class Agency < ActiveRecord::Base
  has_many :offers
end

当我提出这样的要求时 - 一切都很好

@offers = Offer.with_state(:confirmed).
  includes(:destination, :cruise_line, :ship).
  paginate(per_page: 10, page: params[:page]).decorate

但是我想只选择属于active代理商的商品（ agencies表中的列state ），所以我尝试这样做：

@offers = Offer.with_state(:confirmed).
  includes(:destination, :cruise_line, :ship).
  joins(:agency).
  where(agency: {state: 'active'}).
  paginate(per_page: 10, page: params[:page]).decorate

完成这个后，我得到错误PG::UndefinedTable: ERROR: missing FROM-clause entry for table "agency" 。 我的代码出了什么问题？

查询给了我这个错误和sql：

PG::UndefinedTable: ERROR: missing FROM-clause entry for table "agency" LINE 1: ...id" WHERE ("offers"."state" IN ('confirmed')) AND "agency"."... ^ : 
SELECT "offers"."id" AS t0_r0, "offers"."name" AS t0_r1, "offers"."destination_id" AS t0_r2, "offers"."cruise_line_id" AS t0_r3, "offers"."ship_id" AS t0_r4, "offers"."departure_date" AS t0_r5, "offers"."departure_port_id" AS t0_r6, "offers"."arrival_date" AS t0_r7, "offers"."arrival_port_id" AS t0_r8, "offers"."flight_price" AS t0_r9, "offers"."bonus" AS t0_r10, "offers"."itinerary" AS t0_r11, "offers"."board_language_id" AS t0_r12, "offers"."agency_landing_page" AS t0_r13, "offers"."benefits" AS t0_r14, "offers"."inner_price" AS t0_r15, "offers"."inner_price_normal" AS t0_r16, "offers"."outer_price" AS t0_r17, "offers"."outer_price_normal" AS t0_r18, "offers"."balcony_price" AS t0_r19, "offers"."balcony_price_normal" AS t0_r20, "offers"."suite_price" AS t0_r21, "offers"."suite_price_normal" AS t0_r22, "offers"."lucky_price" AS t0_r23, "offers"."lucky_price_normal" AS t0_r24, "offers"."valid_from" AS t0_r25, "offers"."valid_till" AS t0_r26, "offers"."created_at" AS t0_r27, "offers"."updated_at" AS t0_r28, "offers"."description" AS t0_r29, "offers"."agency_id" AS t0_r30, "offers"."state" AS t0_r31, "destinations"."id" AS t1_r0, "destinations"."name" AS t1_r1, "destinations"."created_at" AS t1_r2, "destinations"."updated_at" AS t1_r3, "cruise_lines"."id" AS t2_r0, "cruise_lines"."name" AS t2_r1, "cruise_lines"."created_at" AS t2_r2, "cruise_lines"."updated_at" AS t2_r3, "ships"."id" AS t3_r0, "ships"."name" AS t3_r1, "ships"."picture" AS t3_r2, "ships"."cruise_line_id" AS t3_r3, "ships"."created_at" AS t3_r4, "ships"."updated_at" AS t3_r5 
FROM "offers" 
INNER JOIN "agencies" ON "agencies"."id" = "offers"."agency_id" 
LEFT OUTER JOIN "destinations" ON "destinations"."id" = "offers"."destination_id" 
LEFT OUTER JOIN "cruise_lines" ON "cruise_lines"."id" = "offers"."cruise_line_id" 
LEFT OUTER JOIN "ships" ON "ships"."id" = "offers"."ship_id" 
WHERE ("offers"."state" IN ('confirmed')) AND "agency"."state" = 'active' 
LIMIT 10 OFFSET 0

Answer 1

你唯一添加的是where -clause。 似乎有问题：

where(agency: {state: 'active'})

...并且产生错误的条件，因为在这种情况下，散列键应该表示表名。 你可能应该这样：

where(agencies: {state: 'active'})

更新

看到这一点吸引了很多关注，我想我也应该提出一种不同的方式来做同样的事情，稍微更具组合性：

merge( Agency.where(state: 'active') )

这个怎么样更好？ 它没有假设模型的表名（不是大多数时候都很重要），并允许你使用范围 ：

# Inside Agency
scope :active, -> { where(state: 'active') }

# Somewhere else
merge(Agency.active)